ENERGY CONVERSION ONE (Course 25741)

1
ENERGY CONVERSION ONE (Course 25741)

• Chapter Two
• TRANSFORMERS
• continued

2
Equivalent Circuit of Transformer

• Major Items to be considered in Construction of
  Transformer Model
• Copper losses (in primary Secondary winding)
  I²
• Eddy current losses (in core) V²
• Hysteresis losses (in core) a complex nonlinear
  function of applied V
• Leakage flux fLP fLS, these fluxes
  produce
• self-inductance in primary secondary coils

3
Equivalent Circuit of Transformer

• Exact Eq. cct. Model for Real Transformer
• Copper losses modeled by resistances Rp Rs
• As discussed before
• fpfmfLp fp total av. Primary flux
• fSfmfLS fS total av. Secondary flux
• where fm flux linking both P S
• fLp primary leakage flux
• fLS secondary leakage flux
• The average primary ( Secondary) flux, each,
  is divided into two components as
• mutual flux leakage flux

4
Equivalent Circuit of Transformer

• Based on application of these components,
  Faradays law for primary circuit can be
  expressed as
• Vp(t)Np dfp/dt Np dfM/dt Np dfLp/dt or
• Vp(t)ep(t) eLp(t) similarly for secondary
• Vs(t)Ns dfs/dt Ns dfM/dt Ns dfLs/dt or
• Vs(t)es(t) eLs(t)
• primary secondary voltages due to mutual flux
  
• ep(t) Np dfM/dt es(t) Ns
  dfM/dt

5
Equivalent Circuit of Transformer

• Note ep(t)/Np dfM/dt es(t)/Ns
• ep(t)/es(t) Np / Ns a
• while eLp(t) Np dfLp/dt eLs(t) Ns dfLs/dt
• if ? permeance of leakage flux path
• fLp(p Np) ip fLs(p Ns) is
• eLp(t) Np d/dt (p Np) ip Np²p dip/dt
• eLs(t) Ns d/dt (p Ns) is Ns²p dis/dt
• DefiningLp Np²p primary leakage inductanc
• Ls Ns²p secondary leakage
  inductance

6
Equivalent Circuit of Transformer

• eLp(t)Lp dip/dt
• eLs(t)Ls dis/dt
• Therefore leakage flux can be modeled by primary
  secondary leakage inductances in equivalent
  electric circuit
• Core Excitation that is related to the flux
  linking both windings (fm flux linking both P
  S) should also be realized in modeling
• im (in unsaturated region) e (voltage applied
  to core)
• and lag applied voltage by 90? modeled by an
  inductance Lm (reactance Xm)
• Core-loss current ieh is voltage applied It
  can be modeled by a resistance Rc across primary
  voltage source
• Note these currents nonlinear therefore Xm
  Rc are best approximation of real excitation

7
Equivalent Circuit of Transformer

• The resulted equivalent circuit is shown
• Voltage applied to core input voltage-internal
  voltage drops of winding

8
Equivalent Circuit of Transformer

• to analyze practical circuits including
  Transformers, it is required to have equivalent
  cct. at a single voltage
• Therefore circuit can be referred either to its
  primary side or secondary side as shown

9
Equivalent Circuit of Transformer

• Approximate Equivalent Circuits of a Transformer
• in practice in some studies these models are more
  complex than necessary
• i.e. the excitation branch add another node to
  circuit, while in steady state study, current of
  this branch is negligible
• And cause negligible voltage drop in Rp Xp
• Therefore approximate eq. model offered as

10
Equivalent Circuit of TransformerApproximate
transformer models

• a- referred to primary
• b- referred to secondary
• c- with no excitation branch referred to p
• d- with no excitation branch referred to s

11
Determination of Transformer Eq. cct. parameters

• Approximation of inductances resistances
  obtained by two tests
• open circuit test short circuit test
• 1- open circuit test transformers secondary
  winding is open circuited, primary connected to
  a full-rated line voltage, Open-circuit test
  connections as below

12
Determination of Transformer Eq. cct. parameters

• Input current, input voltage input power
  measured
• From these can determine p.f., input current, and
  consequently both magnitude angle of excitation
  impedance (RC, and XM)
• First determining related admittance and
  Susceptance
• GC1/RC BM1/XM ? YEGC-jBM1/RC -1/XM
• Magnitude of excitation admittance referred to
  primary circuit YE IOC/VOC
• P.f. used to determine angle,
• PFcos?POC/VOC . IOC
• ?cos?1 POC/VOC . IOC

13
Determination of Transformer Eq. cct. parameters

• Thus YE IOC/VOC IOC / VOC
  
• Using these equations RC XM can be
• determined from O.C. measurement

14
Determination of Transformer Eq. cct. parameters

• Short-Circuit test the secondary terminals of
  transformer are short circuited, and primary
  terminals connected to a low voltage source
• Input voltage adjusted until current in s.c.
  windings equal to its rated value

15
Determination of Transformer Eq. cct. parameters

• The input voltage, current and power are again
  measured
• Since input voltage is so low during
  short-circuited test, negligible current flows
  through excitation branch
• Therefore, voltage drop in transformer attributed
  to series elements
• Magnitude of series impedances referred to
  primary side of transformer is ZSE VSC/
  ISC , PFcos?PSC/VSC ISC
• ?cos?1 PSC/VSC ISC
• ZSE VSC / ISC VSC/ ISC
• series impedance ZSE is equal to
• ZSEReqjXeq (RP a²RS) j(XPa²XS)
• It is possible to determine the total series
  impedance referred to primary side , however
  difficult to split series impedance into primary
  secondary components although it is not
  necessary to solve problem
• These same tests may also be performed on
  secondary side of transformer

16
Determination of Transformer Eq. cct. parameters

• Determine Equivalent cct. Impedances of a 20 kVA,
  8000/240 V, 60 Hz transformer
• O.C. S.C. measurements shown?
• P.F. in O.C. is
• PFcos?POC/VOCIOC
• 400 W/ 8000V x 0.214A
• 0.234 lagging

O.C. test (on primary) S.C. test (on primary)
VOC8000V VSC489V
IOC0.214A ISC2.5 A
POC400W PSC240W
17
Determination of Transformer Eq. cct. parameters

• excitation impedance
• YEIOC/VOC
  
• 0.214 A / 8000 V
• 0.0000268
• 0.0000063 j 0.0000261 1/RC- j 1/XM
• Therefore
• RC1/ 0.0000063 159 kO
• XM 1/0.000026138.4 kO

18
Determination of Transformer Eq. cct. parameters

• PF in sc test
• PFcos? PSC/VSCISC240W/ 489x2.50.196
  lagging
• Series impedance
• ZSEVSC/ISC
  
• 489 V/ 2.5 A 195.6
  
• 38.4 j 192 O
• The Eq. resistance reactance are
• Req38.4 O , Xeq192 O

19
Determination of Transformer Eq. cct. parameters

• The resulting Eq. circuit is shown below

20
The Per Unit System For Modeling

• As seen in last Example, solving cct. containing
  transformers requires tedious operation to refer
  all voltages to a common level
• In another approach, the need mentioned above is
  eliminated impedance transformation is avoided
• That method is known as per-unit system of
  measurement
• there is also another advantage, in application
  of per-unit as size of machinery Transformer
  varies its internal impedances vary widely, thus
  a 0.1 O cct. Impedance may not be adequate
  depends on devices voltage and power ratings

21
The Per Unit System For Modeling

• In per unit system the voltages, currents,
  powers, impedance and other electrical quantities
  not measured in SI units system
• However it is measured and define as a decimal
  fraction of some base level
• Any quantity can be expressed on pu basis
• Quantity in p.u.
• Actual Value / base value of quantity

22
The Per Unit System For Modeling

• Two base quantities selected other base
  quantities can be determined from them
• Usually voltage, power
• Pbase,Qbase, or Sbase Vbase Ibase
• Zbase Vbase/Ibase
• YbaseIbase/Vbase
• Zbase(Vbase)² / Sbase
• In a power system, bases for power voltage
  selected at a specific point, power base remain
  constant, while voltage base changes at every
  transformer

23
The Per Unit System For Modeling

• Example A simple power system shown in Figure
  below
• Contains a 480 V generator connected to an ideal
  110 step up transformer, a transmission line, an
  ideal 201 step-down transformer, and a load

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The Per Unit System For Modeling Example

• Impedance of line 20j60O,impedance of load
• Base values chosen as 480 V and 10 kVA at
  genertor
• (a) Find bas voltage, current, impedance, and
  power
• at every point in power system
• (b) convert this system to its p.u.
  equivalent cct.
• (c) Find power supplied to load in this
  system
• (d) Find power lost in transmission line
• (a) At generator IbaseSbase/Vbase
  110000/48020.83 A
• Zbase1Vbase1/Ibase1480
  /20.8323.04 O
• Turn ratio of transformer T1 , a1/10 0.1 so
  base voltage at line Vbase2Vbase1/a480/0.14800
  V

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The Per Unit System For Modeling Example

• Sbase210 kVA
• Ibase210000/48002.083 A
• Zbase24800 V/ 2.083 A 2304 O
• Turn ratio of transformer T2 is a20/120, so
  voltage base at load is
• Vbase3Vbase2/a 4800/20 240 V
• Other base quantities are
• Sbase310 kVA
• Ibase310000/24041.67 A
• Zbase3240/41.67 5.76 O

26
The Per Unit System For Modeling Example

• (b) to build the pu equivalent cct. Of power
  system, each cct parameter divided by its base
  value
• VG,pu
• Zline,pu(20j60)/23040.0087j0.0260 pu
• Zload,pu
• Per unit equivalent cct of PWR. SYS. Shown below

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The Per Unit System For Modeling Example

• (c) current flowing in
• IpuVpu/Ztot,pu
• 
  pu
• Per unit power of load
• Pload,pu Ipu²Rpu(0.569)²(1.503)0.487
• actual power supplied to load
• PloadPload,puSbase0.487 x 100004870 W
• (d) power loss in line
• Pline loss,pu Ipu²Rline,pu(0.569)²(0.0087)0.002
  82
• Pline Pline loss,pu Sbase (0.00282)(10000)28.2
  W